Task Sets: Solve systems of linear equations - Common Core Math - HS Algebra

Concepts

Two linear equations produce the following when solved by elimination:
$0 = 0$ .

What does this result mean about the solution set of the system?

The system has exactly one solution.

The system has no solution (inconsistent).

The system has infinitely many solutions (consistent dependent).

An arithmetic error was made; the result should not be $0 = 0$ .

A student solves the system and finds the solution candidate $(2, -1)$ .
They check: $3(2) + 2(-1) = 4$ ✓ (satisfies the first equation).
They do not check the second equation.

Why is the student's check insufficient?

Checking in the first equation is sufficient — a solution to any one
equation is a solution to the system.

A system requires satisfying ALL equations simultaneously. The student
must verify $(2, -1)$ in both equations to confirm it is a solution.

The check is insufficient because the student should substitute into
the modified equation from elimination, not the original.

Graphical verification is required; algebraic checking is not accepted.

Procedures

Solve the system using substitution:
$y = 2x - 3$
$3x + y = 7$

Substitute the first equation into the second:
$3x + (2x - 3) = 7 \Rightarrow 5x = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $\Rightarrow x = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

Then $y = 2($ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $) - 3 = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

sum-after-combine:

x-value:

x-again:

y-value:

Solve the system using elimination:
$2x + 3y = 12$
$2x - y = 4$

Subtract the second equation from the first to eliminate $x$ :
$(2x + 3y) - (2x - y) = 12 - 4 \Rightarrow 4y = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $\Rightarrow y = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

Then $2x - ($ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $) = 4 \Rightarrow x = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

rhs:

y-value:

y-sub:

x-value:

After using elimination on a system, a student finds $y = 3$ . They
now need to find $x$ .

The original system was:
$E_1: 5x - 2y = 4$
$E_2: 3x + 4y = 30$

During elimination, the student multiplied $E_1$ by 2 to get $E_1': 10x - 4y = 8$ .
They then added $E_1'$ and $E_2$ to get $13x = 38$ , giving $x = \frac{38}{13}$ .

Where should the student substitute $y = 3$ to find $x$ ?

Into the modified equation $E_1'$ : $10x - 4y = 8$

Into $13x = 38$ (the equation obtained by adding $E_1'$ and $E_2$ )

Into either original equation ( $E_1$ or $E_2$ )

Into the average of $E_1$ and $E_2$

HSA.REI.C.6 Tutor Intake — Systems of Linear Equations

Concepts

Procedures